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I want to know if it is possible to solve a QCQP problem with quadratic equality constraints in SDP. I know it is possible to convert a QCQP to an SDP by using the Shur complement. The following worked for me thus far: $$ \begin{equation} \begin{array}{cccccc} \underset{x}{\min} & x^{T}Q_{0}x+q_{0}^{T}x+c_{0} & & \underset{t,x}{min} & t\\ s.t & x^{T}Q_{i}x+q_{i}^{T}x+c_{i}\leq0 & \Longleftrightarrow & s.t & \left(\begin{array}{cc} I & M_{0}x\\ x^{T}M_{0}^{T} & -c_{0}-q_{0}^{T}x+t \end{array}\right) & \succeq0\\ & & & & \left(\begin{array}{cc} I & M_{i}x\\ x^{T}M_{i}^{T} & -c_{i}-q_{i}^{T}x \end{array}\right) & \succeq0\\ & i=1,2,\dots,m & & & i=1,2,\dots,m & \end{array} \end{equation} $$

where $M_{j}M_{j}^{T}=Q_{j}$ (eigendecomposition can also be used, thanks Alt)

However it seems this only applies to quadratic inequality constraints.

I considered converting the desired inequality constraints to equalities by using slack variables but I think that technique only works with linear constraints. I also considered having two constraints like: $$ \begin{equation} \begin{array}{c} x^{T}Qx+q^{T}x+c\leq0\\ -(x^{T}Qx+q^{T}x+c)\leq0 \end{array} \end{equation} $$ To force an equality but the second constraint is not convex so it wont work.

So is there another way to do it?

Edit: turns out quadratic equality constraints are non-convex thus cant be solved directly with this approach

Rodrigo de Azevedo
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Kirillvh
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2 Answers2

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Hint:

You don't need the constraints $M_{j}M_{j}^{T}=Q_{j}$. For example if $UDU^T$ is the eigen decomposition of $Q_j$, Then $M_j=UD^{\frac{1}{2}}$ (if $Q_j$ is symmetric)

Note that if you introduce new variable $M_j$, then the bilinear forms in the constraints make your problem non-convex, (Besides the fact that equality constraints with quadratic form $M_{j}M_{j}^{T}=Q_{j}$that you have to add also makes the problem non-convex.)

Alt
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  • Thank you for the hint. Using the eigen decomposition instead of Cholesky might allow me to Decompose $-Q_i$, problem was that $Q_i$ was positive definite so $-Q_i$ was negative definite and Cholseky cant decompose that. However to clarify though, I dont mean that $M_jM_j^T=Q_j$ is a constraint but rather a once off operation I perform when converting from the QCQP to the SDP (I will edit to clarify). – Kirillvh Apr 30 '14 at 07:34

  • equality constraints with quadratic form $M_jM^T_j=Q_j$ that you have to add also makes the problem non-convex.

    Thank you, with that hint and this I realized that quadratic equality constraints are non-convex thus cant be solved directly, but perhaps relaxation technique can be used to solve such a problem.

     – Kirillvh Apr 30 '14 at 10:10
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I cannot see why you should convert a QCQP in an SDP problem. You may want instead to move to a SOCP formulation, which is usually more robust and enjoy a simpler duality theory.

Solvers for QCQP are far more efficient and reliable than SDP ones.

AndreaCassioli
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    Yes you are right, but I didnt ask about the merits/demerits of SDP. – Kirillvh Apr 30 '14 at 07:39
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    I was just wondering the reason, out of my curiosity. – AndreaCassioli Apr 30 '14 at 12:35
  • By the way, a quadratic equality is not convex by definition...if I am not wrong you can only get a convex SDP relaxation. I suggest you to search for the sensor network localization problem, there are some SDP relaxations of contraints of the form ||x-y||^2 = z that might be useful. – AndreaCassioli Apr 30 '14 at 21:53